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2. How much water at 0 per gallon, must be mixed with wine at 90 cents per gallon, so as to fill a vessel of 100 gallons, which may be afforded at 60 cents per gallon ?
Ans. 33} gals. water, and 66; gals. wine. 3. A grccer having sugars at 8 cts. 16 cts. and 24 cts. per pound, would make a composition of 240 lb. worth 20 cts. per Ib. without gain or loss; what quantity of each must be taken ?
Ans. 40 lb. at 8 cts. 40 at 16 cts, and 160 at 24 cts. 4. A goldsmith had two sorts of silver bullion, one of 10 oz. and the other of 5 07., fine, and has a mind to mix a pound of it so that it shall be 8 oz fine; how much of each sort must he take?
fins, 44 of 5 oz. fine, and 7 of 10 oz. fine. 5. Brandy at 3s. Od. and 5s. 9d. per gallon, is to be mixed, so that a hhd. of 6s gallons may be sold for 121. 125.; how many gallons must be taken of each?
Ans. 14 gals. at 5s. 9d. and 49 gals. at Ss. 6d.
ARITHMETICAL PROGRESSION. ANY rank of numbers more than two, increasing by common excess, or decreasing by common difference, is said to be in Arithmetical Progression.
52, 4, 6, 8, &c. is an ascending arithmetical series :
28, 6, 4, 2, &c. is a descending arithmetical series : The numbers which form the series, are called the terms of the progression; the first and last terms of which are called the extremes.
PROBLEM I. The first term, the last term, and the number of terms being given, to find the sum of all the terms.
* A series in progression includes five parts, viz. the first term, last term, number of terms, common difference, and sum of the series.
By having any three of these parts given, the other two may be found, which admits of a variety of Problems; but most of them are best understood by an algebraic process, and are here omitted.
RULE. Multiply the sum of the extremes by the number of terms, and half the product will be the answer.
1. The first terın of an arithmetical series is 3, the last term 25, and the number of terms 11; required the sum of the series.
23+3=26 sum of the extremes.
Then 26x11:22:145 the Answer. 2. How many strokes does the hammer of a clock trike, in twelve hours ?
Ans. 78. 3. A merchant sold 100 yards of cloth, viz. the first yard for 1 ct. the second for: cts. the third for 3 cts. &c. I demand what the cloth came to at that rate ?
Ans. $501. 4. A man bought 19 yards of linen in arithmetical progression, for the first yard he gave 1s. and for the last yd. il. 17s. what did the whole coine to ? Ans. £18 1s.
5. A draper sold 100 yards of broadcloth, at 5 cts. for the first yard, 10 cts. for the second, 15 for the third, &c. increasing 5 cents for every yard ; what did the whole amount to, and what did it average per yard ?
Ins. Amount 8952), and the average price is $2, 52cts. 5 mills per yard.
6. Suppose 144 oranges were laid 2 yards distant from each other, in a right line, and a basket placed two yards from the first orange, what length of ground will that boy travel over, who gathers them up singly, returning with them one by one to the basket ?
Ans. 23 miles, 5 furlongs, 180 yds,
PROBLEM II. The first term, the last term, and the number of terms given, in find the common difference.
RULE. Divide the difference of the extremes by the number of terms less 1, and the quotient will be the common dif: ference.
1. The extremes are 3 and 29, and the number a terms 14, what is the common difference:
-SS Number of terms less 15-13)26(2 Ans. 2. A man had 9.sons, whose several
differed alike. he youngest was 3 years old, and the oldest 35; wha. was the common difference of their
Ans. 4 years 3. A man is to travel from New-London to a certain place in 9 days, and to go but 3 miles the first day, in. ercasing every day by an equal excess, so that the last clay's journey may be 43 miles : Required the daily in. crease, and the length of the whole journey?
Ans. The daily increase is 5, and the whole journey 207 miles.
4. A debt is to be discharged at 16 different paymeru (in arithmetical progression, the first payment is to in 141. the last 100l.: What is the common difference, and the sum of the whole debt?
Ans. 5l. 14s. 8d. cominon difference, and 9121. the whole debi,
PROBLEM III. Given the first term, last term, and common difference to
find the number of terms.
RULE. Divide the difference of the extremes by the common difference, and the quotient increased by i is the number of terms.
1. Jf the extremes be 3 and 45, an' the common dif serence %; what is the number of terms ? Ans. 22.
2. A man going a journey, travelled the first day five miles, the last day 45 miles, and cach day increased his Journry by 4 miles; how many days did he travel and how far: a ns. 11 days, and the whole distance, travelled 275 miles
GEOMETRICAL PROGRESSION, Is when any rank or series of numbers increased by one common multiplier, or decreases by one common divisor ; à as 1, 2, 4, 8, 10, &c. increase by the multiplier 2; and 27, 9, 3, 1, decrease by the divisor 3.
PROBLEM I. The first term, the last term (or the extremes) and the ratio given, to find the sum of the serieś.
RULE. Multiply the last term by the ratio, and from the pro duct subtract the first term ; then divide the remainder by the ratio, less by 1, and the quotient will be the sum. of all the terms.
1. If the series be 2, 6, 18, 54, 162, 486, 1458, and the ratio 3, what is its sum total ?
=-2186 the Answer.
S-1 2. The extremes of a geometrical series are 1 and 65536, and the ratio 4 ; what is the sum of the series ?
Ans, 87381. PROBLEM II. Given the first term, and the ratio, to find any other term
CASE I. When the first term of the series and the ratio are equal.
*As the last term in a long series of numbers is very te. dious to be found by continual multiplications, it will be necessary for the readier finding it out, to have a series of numbers in arithmetical proportion, called indices, whose common difference is 1.
When the firsi term of the series and the ratio are equal, the indices must begin with the unit, and in this ease, the
1. Write down a few of the leading terms of the series, and place their indices over them, beginning the indices with an unit or 1.
2. Add together such indices, whose sum shall make up the entire index to the sum required.
3. Multiply the terms of the geometrical series belonging to those indices together, and the product will be the term sought.
1. If the first be 2, and the ratio %; what is the 13th terin. 1, 2, 3, 4, 5, indices. Then 5+5+3=13 2, 4, 8, 16, 32, leading terins. 32x32x8=8192 Ans.
2. A draper sold 20 yards of superfine cloth, the first yard for 3d. the second for 9d. the third for 27d. &c. in triple proportion geometrical ; what did the cloth come to at that rate ?
The 20th, or last term is $486784401d. Then 3+3486784401.-3
5230176600d. the sum of all
3-1 the terms (by Prob. I.) equal to £21792402 10s. Ans.
S. A rich miser thought 20 guineas a price too much for 12 fine horses, but agreed to give 4 cents for the first, 16 cents for the second, and 64 cents for the third horse, and so on in quadruple or fourfold proportion to the last: what did they come to at that rate, and how much did they cost per head, one with another ?
Ans. The 12 horses came to $223696, 20cts. and the average price was 818641, 35cts, per head.
product of any two terms is equal to that term, signified by the sum of their indices.
$1 2 3 4 5 &c. Indices or arithmetical seria. Thus, 12 4 8 16 32 &c. geometrical series. Now,
3+2 5 = the index of the fifth terms and 4*8 = 32 = the fifth term.